Designing Capable and Reliable Products Episode 2 Part 3 docx

Determining the variables For an adequate interference ®t selected on the basis of the hole, the tolerances are taken from BS 4500A (1970) as H7-s6. This translates to a tolerance on the hub bore H7 0:030, 0:000 mm and for the shaft diameter, s6 0:072, 0:053 mm. Given the notation for hub bore diameter is d H , and converting to mean values and bilateral tolerances, these dimensions become: d H  150:015 Æ 0:015 mm d S % 150:063 Æ 0:010 mm and from Figure 4.55, D H  170 Æ 0:05 mm l H  100 Æ 0:05 mm The hub inside diameter, outside diameter and length are machined using a lathe and so we employ the turning/boring map as shown previously in Figure 4.42. The material speci®ed for the hub is mild steel giving a material to process risk m p  1:3. The geometry to process risk g p  1:02 due to a 2 X 1 length to diameter ratio. An adjusted tolerance for the hub bore, d H , is then given by: Adjusted tolerance  Design tolerance t m p  g p  0:015 1:3  1:02  0:011 With reference to the process capability map for turning/boring, A  1:7 for a dimen- sion of 150 mm. This value defaults to the component manufacturing variability risk, q m , when there is no consideration of surface ®nish capability in an analysis à . The shifted standard deviation,  H , for the dimensional tolerance on the hub bore can then be predicted from equation 4.28:  H d H  t Á q 2 m 12  0:015  1:7 2 12  0:004 mm Therefore: d H $ N50:015; 0:004mm Similarly for an analysis on the hub outside diameter and length, which are turned, and the shaft diameter, which is ®nished using cylindrical grinding, we get: D H $ N70; 0:005mm l H $ N100; 0:006mm d S $ N50:063; 0:0012mm à Note that the surface roughness to process risk, s p , is not included in the formulation to determine the standard deviation of the tolerance. However, a 0.4 mm Ra surface ®nish is required to provide adequate frictional adhesion between the shaft and hub bore (Bolz, 1981). This capability requirement is met by the manufacturing processes selected when analysed using CA. Case studies 225 The mean of the interference between the shaft and hub bore is given by:     d S ÿ  d H  50:063 ÿ 50:015  0:048 mm The standard deviation of the interference is given by:     H2 d S   H2 d H ÿÁ 0:5  0:0012 2  0:004 2 ÿÁ 0:5  0:0042 mm Therefore,  $ N0:048; 0:0042mm The maximum coecient of variation for the Modulus of Elasticity, E, for carbon steel was given in Table 4.5 as C v  0:03. Typically, E  208 GPa and therefore we can infer that E is represented by a Normal distribution with parameters: E $ N208; 6:24GPa Values typically range from 0.077 to 0.33 for the static coecient of friction for steel on steel under an interference ®t with no lubrication (Kutz, 1986). The interference and coecient of friction are correlated in practice but for the example here, we assume statistical independence. Also, assuming that 6 standard deviations cover the range given, we can derive that: f $ N0:2; 0:04 Determining the probability of interference The mean value for the holding torque capacity,  M H , is found by inserting the mean values of all the variables into equation 4.87. To ®nd the standard deviation of the holding torque,  M H , we can apply the Finite Dierence Method. Finally, we arrive at: M H $ N3:84; 0:85kNm At this stage it is worth highlighting the relative contribution of each variable to the holding torque variance. The results from the Finite Dierence Method are used to construct Figure 4.56 which shows the sensitivity analysis of all the variables to the variance of the holding torque. It is clear that accurate and representative data for the coecient of friction, f, for a particular situation and the control of the interfer- ence ®t dimensions are crucial in the determination of the holding torque distribution. The torque that can be transmitted by the shaft without yielding, M S , is given by: M S   y Á J r where:  y  shear yield strength J  polar second moment of area r  radius of shaft: In terms of the shaft diameter, this simpli®es to: M S  0:1963495 y Á d 3 S 4:88 226 Designing reliable products The shear yield strength for ductile metals is a linear function of the uniaxial yield strength. Therefore, for pure torsion from equation 4.56:  y  0:577Sy Applying this conversion to the Normal distribution parameters for SAE 1035 steel gives:  y $ N197:3; 15:2MPa Using Monte Carlo simulation applied to equation 4.88, the shaft torque capacity is found to be: M S $ N4:86; 0:38kNm Both of the torque capacities calculated, the holding torque of the hub and the shaft torque at yield, are represented by the Normal distribution, therefore we can use the coupling equation to determine the probability of interference, where: z ÿ  M S ÿ  M H   2 M S   2 M H q ÿ 4:86 ÿ 3:84  0:38 2  0:85 2 p ÿ1:10 From Table 1 in Appendix I, the probability of failure P  0:135666. This suggests that when the holding torque level is reached in service, due to a malfunction stopping the hub from rotating, there is about a 1 in 7 chance that the shaft will yield before the hub slips. Clearly this is not adequate if we want to protect the shaft and its transmis- sion from damage. Rather than change the speci®cations of the design as a whole, a small increase in the yield strength of the shaft is favoured. This is because there are only two variables in the shaft strength and it is obvious that the major contribution to the torque is from the shear yield strength. Repeating the calculation to determine the shaft torque at yield using cold drawn SAE 1018 with Sy $ N540; 41MPa gives P  0:000104 or 0 10 20 30 40 50 60 70 80 90 Variance contribution (%) Design variable Figure 4.56 Sensitivity analysis for the holding torque variables Case studies 227 a reliability R  0:999896. Although not safety critical, the degree of protection from the shaft yielding is now adequate after reference to the reliability target map. The relative shape of the torque distributions and degree of interference are shown in Figure 4.57 for the two situations where dierent shaft steels are used. 4.8.4 Weak link design The major assumption in weak link design is that the cost of failure of the machine that is to be protected from an overload situation in service is much greater than the cost of failure of a weak link placed in the system which is designed to fail ®rst. The situation is primarily driven by various costs which must be balanced to avoid at one extreme the cost of failure of the system, and the other overdesign of the elements in the system. The cost factors involved are typically: . Cost of introducing the weak link . Cost of replacement of the weak link Figure 4.57 Distributions and relative interference of the holding torque and shaft torque capacities for two different shaft steels 228 Designing reliable products . Cost of failure with the weak link relative to the cost of failure without the weak link . Cost due to machine downtime if the weak link fails prematurely . Cost of increasing strength of machine elements to accommodate the weak link. Figure 4.58 shows the concept of weak link design. The loading stress distribution is determined from the normal operating conditions found in the system and this stress is used to determine the dimensions of the weak link. The failure mode for the weak link is stress rupture and so the ultimate tensile or ultimate shear strength is speci®ed. The appropriate level of interference between the loading stress and weak link strength is determined from the consequences of machine downtime if the weak link fails prematurely. The use of the target reliability map with FMEA Severity Ratings (S) for production processes is useful in this respect (see the Process FMEA Severity Ratings provided by Chrysler Corporation et al., 1995). For example, in Section 2.6.4, the characteristic dimension on the cover support leg was critical to the success of the automated assembly process, the potential failure mode being a major disruption to the production line on failure to meet the required capability, therefore S  8 was allocated. If the weak link was designed to a very low level of failure probability, meaning the separation between the two distributions was greater, the machine elements could be overdesigned. The variability of the weak link strength should be well known because it is a critical component in the system and experimental testing to determine the ultimate tensile strength, Su, is recommended where data is lacking. The interference between the weak link strength and machine strength is usually always smaller comparatively than that between the loading stress and weak link strength. This is due to the fact that we must see the failure of the weak link before the machine in all situations. Figure 4.58 The weak link concept Case studies 229 However, too great a separation, and overdesign may occur. The overload condition is represented by a unique stress, which is very much greater than the working stress, applied suddenly which causes only the weak link to failure due to stress rupture. In the following case study we will examine the concepts of weak link design. Figure 4.59 shows the arrangement of a coupling with a shear pin acting as a weak link between a transmission system and a pump, the assumption being that the cost of failure of the pump is much greater than the cost of failure of the weak link. In the event that the pump suddenly stops due to a blockage, the shear pin must protect the system from damage. The applied torque through the transmission shaft under normal running conditions is M  3:2 kNm with a coecient of variation C v  0:1, and the diameter variables of the transmission and pump shafts are D $ N60; 0:004mm. The FMEA Severity Rating S5, relating to a minor disruption if the weak link fails prematurely causing the pump to experience down- time. Steel is to be used as the material for all the machine elements. Experimental determination of ultimate tensile strength of the weak link material Because the design of the shear pin is critical, the ultimate tensile strength of the steel selected for the weak link material was in this case measured statistically by perform- ing a simple experimental hardness test. The grade of steel selected is 220M07 cold drawn free cutting steel. The size tested is 116 mm, estimated as the approximate diameter of the pin, and 30 samples are selected from the stock material. The Brinnel Hardness (HB) value of each sample is measured, the results of which are shown in Table 4.14. Rather than develop a histogram for the data, we can determine the Normal distribution plotting positions using the mean rank equation for the indivi- dual values. The results are plotted in Figure 4.60, the equation of the straight line and the correlation coecient, r, determined. The mean and standard deviation of the hardness for the steel can be determined from the regression constants A0andA1 as:  HB ÿ  A0 A1  ÿ  ÿ34:801 0:2319   150:07  HB   1 ÿ A0 A1    A0 A1    1  34:801 0:2319    ÿ34:801 0:2319   4:31 Figure 4.59 Weak link arrangement 230 Designing reliable products Table 4.14 Analysis of hardness data and plotting positions for the Normal distribution HB (x-axis) Cum. freq. (i) (N  30) F i  i N  1 z  È ÿ1 SND F i  ( y-axis) 141.8 1 0.0323 ÿ1.849 144.5 2 0.0645 ÿ1.518 145.4 3 0.0968 ÿ1.300 145.4 4 0.1290 ÿ1.131 145.4 5 0.1613 ÿ0.989 145.9 6 0.1935 ÿ0.865 147.8 7 0.2258 ÿ0.753 147.8 8 0.2581 ÿ0.649 147.8 9 0.2903 ÿ0.552 147.8 10 0.3226 ÿ0.460 147.8 11 0.3548 ÿ0.372 149.3 12 0.3871 ÿ0.287 150.3 13 0.4192 ÿ0.204 150.3 14 0.4516 ÿ0.121 150.3 15 0.4839 ÿ0.041 150.3 16 0.5161 0.041 150.3 17 0.5484 0.121 150.3 18 0.5806 0.204 150.3 19 0.6129 0.287 150.3 20 0.6452 0.372 151.3 21 0.6774 0.460 152.8 22 0.7097 0.552 152.8 23 0.7419 0.649 152.8 24 0.7742 0.753 152.8 25 0.8065 0.865 154.4 26 0.8387 0.989 155.4 27 0.8710 1.131 156.5 28 0.9032 1.300 157.0 29 0.9355 1.518 158.1 30 0.9677 1.849 Figure 4.60 Linear regression for the Normal distribution transformed hardness data Case studies 231 From equations 4.12 and 4.13, the mean and standard deviation for the ultimate tensile strength, Su, for steel can be derived:  Su  3:45 HB  3:45  150:07  517:7 MPa  Su 3:45 2 Á  2 HB  0:152 2 Á  2 HB  0:152 2 Á  2 HB  0:5 3:45 2  4:31 2  0:152 2  150:07 2  0:152 2  4:31 2  0:5  Su  27:2 MPa Therefore: Su $ N517:7; 27:2MPa Typically for ductile steels, the ultimate shear strength,  u is 0.75 of Su (Green, 1992), therefore:  u $ N388:3; 20:4MPa Determining the diameter of the shear pin Assuming that an adequate transition ®t is speci®ed for both the pin hole and coupling bore, the pin is in double pure shear with the bore of the coupling, and so the shear stress, L, is given by: L  F 2A where: F  tangential force A  area of pin: The tangential force acting on the pin at a radius, r, due to the applied torque, M,is given by: F  M r Therefore, combining the above equations and substituting the radius for the dia- meter variable gives: L  1:27324M D Á d 2 4:89 Rearranging equation 4.89 in terms of the pin diameter, d, gives: d   1:27324M D Á L r 4:90 where: M  applied torque D  shaft diameter L  loading stress: 232 Designing reliable products For FMEA S5, the reliability of the weak link in service is required to be R  0:999 with reference to the reliability target map given in Figure 4.36. This relates to a failure probability P  0:001 or when working with the SND, z ÿ3:09 from Table 1 in Appendix I. For a given value of the Standard Normal variate, z, the coupling equation for the interference of two Normal distributions can be used to determine the loading stress, and hence the diameter of the shear pin. From the coupling equation we get: 3:09   u WL ÿ  L   2 u WL   2 L q 4:91 We know that the coecient of variation, C v , of the applied torque is approximately 0.1, and that the ®nal loading stress variable will have a similar level of variation because the dimensional variables have a very small variance contribution in compar- ison. We also know the ultimate shear strength parameters of the weak link material, therefore substituting in equation 4.91 and rearranging to set the right-hand side to zero gives: 3:09  20:4 2  0:1 L  2 q   L ÿ 388:3  0 which yields  L % 281 MPa when solved by iteration. From equation 4.90, the mean pin diameter,  d , is found to be:  d   1:27324 Á  M  D Á  L s   1:27324  3200 0:06  281  10 6 r  0:015545 m  15:545 mm The pin is machined and cylindrically ground to size. It can be shown that the Normal distribution parameters of the diameter d $ N15:545; 0:0005mm for a tolerance of Æ0:002 mm chosen from the relevant process capability map. Solving equation 4.89 using Monte Carlo simulation for the variables involved, the shear stress in the pin is found to have a Normal distribution of L $ N281; 28:3MPa. Calculating the reliability using the coupling equation for the given stress and strength parameters gives R  0:998950 which is almost exactly that speci®ed initially. Selecting the pump shaft material The torque capacity of the pump shaft must be greater than the torque capacity of the shear pin in all cases. We assume that failure of the pump shaft occurs at the interfer- ence of these two torque distributions. From equation 4.89, the torque capacity of the shear pin can be determined by substituting the ultimate shear strength of the weak link material,  u WL for L, giving: M WL  0:785398D Á d 2 Á  u WL 4:92 Solving equation 4.92 using Monte Carlo simulation for the variables involved, the torque capacity of the shear pin is found to have a Normal distribution of M WL $ N4421:7; 234:1Nm. The coecient of variation of the yield strength, and hence the shear yield strength for steels, is typically C v  0:08. This means that the coecient of variation of the Case studies 233 torque capacity of the pump shaft will approximately be the same, because the dimen- sional variable of the shaft diameter is very small in comparison to the shear yield strength. We also need to speculate about the likely failure probability acceptable between the weak link and pump shaft torque capacities. Assuming that the overload situation is only likely to occur once in 1000 operating cycles, the coupling equation can be written as: 3:09   M P ÿ  M WL   2 M P   2 M WL q 4:93 Rearranging equation 4.93 to set the right-hand side to zero and substituting in the known parameters gives: 3:09  0:08 M P ÿÁ 2 234:1 2 q  4421:7 ÿ  M P  0 which yields  M P % 6092:5 Nm by iteration. Therefore, the Normal distribution parameters for the pump shaft torque capacity are: M P $ N6092:5; 487:4Nm The torque that can be transmitted by a shaft without yielding was given in equation 4.88. Rearranging for the shear yield strength and the variables in this example gives:  y  5:09296M P D 3 4:94 Solving equation 4.94 using Monte Carlo simulation for the variables involved, the shear yield strength required for the pump shaft material is found to have a Normal distribution with parameters:  y $ N143:7; 11:6MPa Therefore, the Normal distribution parameters for the material's tensile yield strength are 1=0:577 times greater, giving: Sy $ N249; 20:1MPa A suitable material would be hot rolled mild steel 070M20, which has a minimum yield strength, Sy min  215 MPa (BS 970, 1991). By considering that the minimum yield strength is ÿ3 standard deviations from the mean and that the typical coecient of variation C v  0:08 for the yield strength of steel, the Normal distribution parameters for 070M20 can be approximated by: Sy $ N282:9; 22:6MPa From equation 4.94 and using Monte Carlo simulation again, the actual torque capacity of the pump shaft using the 070M20 steel grade is found to be: M P $ N6922:9; 560:1Nm Finally, the interference between the applied torque and the pump shaft torque capacity can be analysed to determine if separation between them is too great leading 234 Designing reliable products [...]... stress parameters and reliability for a range of standard unequal a, b, t (mm) LA (MPa) LA (MPa) LB (MPa) LB (MPa) RA RB 30  60  5 30  60  6 50  65  5 50  65  6 50  65  8 50  75  6 50  75  6 30 5 .2 264 .2 211.1 1 83. 0 148 .2 1 42. 6 114.0 45.1 39 .2 31 .2 27.0 22 .0 21 .0 16.8 34 6.9 29 7.0 26 4.8 22 5.9 ÿ176 .2 ÿ1 72. 2 ÿ 134 .5 50.8 43. 6 38 .8 32 . 9 26 .0 25 .1 19.6 0.648 724 0.918951 0.99 929 9 0.999995... 1.000000 1.000000 1.000000 0 .33 039 1 0.71 531 9 0.918648 0.9961 12 0.999999 1.000000 1.000000 23 9 24 0 Designing reliable products mean value as described by equation 4 . 23 : Symin ˆ Sy ÿ 3 Sy Therefore, 27 5 ˆ Sy ÿ 3 0:05Sy † Sy ˆ 27 5 ˆ 32 3 :5 MPa 0:85 and Sy ˆ 0:05  32 3 :5 ˆ 16 :2 MPa The yield strength of Grade 43C structural steel can be approximated by: Sy $ N… 32 3 :5; 16 :2 MPa The coupling equation... about the principal axes, u±u and v±v, respectively are given by: q ÿ Á ÿ 2 2 Iuu ˆ 1 Ixx ‡ Iyy ‡ 1 Ixx ÿ Iyy ‡4Ixy …4:100† 2 2 q ÿ Á ÿ 2 2 Ixx ÿ Iyy ‡4Ixy …4:101† Ivv ˆ 1 Ixx ‡ Iyy ÿ 1 2 2 23 7 23 8 Designing reliable products where the principal plane is at an angle: tan ˆ ÿ1  2Ixy Iyy ÿ Ixx 2  …4:1 02 The above equations can all... 1:067  105  4  10ÿ6 U T ÿ 2 ÿ 2 ÿ 2 ÿ 2 U T U T ‡ 8:41  104  2  10ÿ6 ‡ 1:1 72  10ÿ11  6 :24  109 T ÿ ÁT ˆ T Á ÿ Á ÿ Á ÿ Á U ÿ11 2 9 2 7 2 ÿ6 2 U U T ‡ 2: 419  10  2: 1  10 ‡ 1:4 72  10  0: 127  10 S R ÿ 2 ÿ 2 ÿ 2 ÿ 2 ‡ ÿ1:485  107  0:185  10ÿ6 ‡ 2: 421  104  1:7  10ÿ5 Case studies and ÁT ˆ 3: 418C and the mean ÁT ˆ 84:748C Therefore, the setpoint temperature is given by: T ˆ ÁT... P Q0:5        @ÁT 2 2 @ÁT 2 2 @ÁT 2 2 @ÁT 2 2 Á ymax ‡ Á d S ‡ Á dB ‡ Á  ES U T @y @dS @dB @ES T U max ÁT % T  U 2  2  2  2 R S @ÁT @ÁT @ÁT @ÁT 2 2 2 2 ‡ Á E B ‡ Á S ‡ Á B ‡ Á ` @EB @S @B @` …4:114† The mean value, ÁT , can be approximated by substituting the mean values for each variable into equation 4.114 All eight variables are assumed to be random in nature following... 0:0000 02 m ES $ N 20 8; 6 :24 † GPa EB $ N…105; 2: 1† GPa S $ N… 12: 7; 0: 127 † 10ÿ6 =8C B $ N…18:5; 0:185† 10ÿ6 =8C ` $ N…0:07; 0:000017† m With reference to Appendix XI, we can solve each partial derivative term in equation 4.114 using the Finite Di€erence method to give: Q0:5 Pÿ 2 ÿ 2 ÿ 2 ÿ 2 2:648  104  1:7  10ÿ5 ‡ 1:067  105  4  10ÿ6 U T ÿ 2 ÿ 2 ÿ 2 ÿ 2 U T U T ‡ 8:41  104  2  10ÿ6... by: " xˆ a2 ‡ …b ÿ t†t 2 a ‡ …b ÿ t†† …4:95† " yˆ b2 ‡ …a ÿ t†t 2 a ‡ …b ÿ t†† …4:96† The second moments of area about x±x and y±y respectively are:  à " " Ixx ˆ 1 t…b ÿ y 3 ‡a "3 ÿ …a ÿ t†…y ÿ t 3 y 3  à " " Iyy ˆ 1 t…a ÿ x 3 ‡b "3 ÿ …b ÿ t†…x ÿ t 3 y 3 …4:97† …4:98† The product moment of area is:   ! h i  a  t t bÿt " " " " Ixy ˆ …a ÿ x† ÿ  yÿ  at ‡ x ÿ  bÿyÿ  …b ÿ t†t 2 2 2 2 …4:99† The... Modulus of Elasticity for brass: …4:1 12 24 1 24 2 Designing reliable products Replacing the change in temperature …T ÿ T0 † with ÁT, which is a random variable itself, and rearranging for this term gives: 4 5  2   dB dB EB dB 3 ES Á dS 2 ymax Á dS Á 4 ‡ 6 ‡ 4 ‡ ‡ dS dS E S dS EB Á dB ÁT ˆ …4:1 13 2 3` Á …S ÿ B †…dB ‡ dS † The variance equation to determine the standard deviation of the change in temperature... around half its current design value ÿ Á ÿ Á 2 ˆ …0 :20 3 ‡ …0:1 82 ‡ …0: 028 † ‡ 5 :34 5  10 3 ‡ 2: 579  10 3 ÁT ‡ 3: 497† ‡ …7:550† ‡ …0:169† ˆ 11: 637 For the ®rst term, the variance contribution of the de¯ection variable, ymax , to the variance of the temperature as a percentage becomes: 0 :20 3  100 ˆ 1:74% 11: 637 Repeating the above for each variable and ranking the percentage values in descending order... approximates to 3 kg The mass of the items conveyed was stated to range from 0 to 72 kg and it was assumed this mass varies randomly according to a Normal distribution Again we assume that this mass range approximates to 6 standard deviations, which gives a mean of 36 kg and a standard deviation of 12 kg The total mass can be represented by a Normal distribution with a mean m ˆ 50 ‡ 36 ˆ 86 kg, and the standard . R A R B 30 Â60  5 30 5 .2 45.1 34 6.9 50.8 0.648 724 0 .33 039 1 30 Â60  6 26 4 .2 39 .2 29 7.0 43. 6 0.918951 0.71 531 9 50 Â65  5 21 1.1 31 .2 26 4.8 38 .8 0.99 929 9 0.918648 50 Â65  6 1 83. 0 27 .0 22 5.9 32 . 9. 0 .3 72 151 .3 21 0.6774 0.460 1 52. 8 22 0.7097 0.5 52 1 52. 8 23 0.7419 0.649 1 52. 8 24 0.77 42 0.7 53 1 52. 8 25 0.8065 0.865 154.4 26 0. 838 7 0.989 155.4 27 0.8710 1. 131 156.5 28 0.90 32 1 .30 0 157.0 29 0. 935 5. derived:  Su  3: 45 HB  3: 45  150:07  517:7 MPa  Su  3: 45 2 Á  2 HB  0:1 52 2 Á  2 HB  0:1 52 2 Á  2 HB  0:5  3: 45 2  4 :31 2  0:1 52 2  150:07 2  0:1 52 2  4 :31 2  0:5  Su  27 :2 MPa Therefore: Su